Let,

$C_1\equiv x^2+y^2-2x+4y+4=0$

Centre, $c_2\equiv \left(1,-2\right)$ and radius $r_1=1$

$C_2\equiv x^2+y^2+4x-2y+1=0$

Centre, $c_1\equiv \left(-2,1\right)$ and radius $r_2=2$

Now, distance between centres is given by

$c_1{c}_2=\sqrt{{\left(-2-1\right)}^2+{\left(1+2\right)}^2}=\sqrt{18}$

Now, we know that the length of transverse common tangent is given by

$\sqrt{{\left(c_1{c}_2\right)}^2-{\left(r_1+r_2\right)}^2}$

$=\sqrt{18-{\left(1+2\right)}^2}=3$.